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FRAMES AND MACHINES Today’s Objectives: Students will be able to:

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Presentation on theme: "FRAMES AND MACHINES Today’s Objectives: Students will be able to:"— Presentation transcript:

1 FRAMES AND MACHINES Today’s Objectives: Students will be able to:
a) Draw the free body diagram of a frame or machine and its members. b) Determine the forces acting at the joints and supports of a frame or machine. In-Class Activities: Check Homework, if any Reading Quiz Applications Analysis of a Frame/Machine Concept Quiz Group Problem Solving Attention Quiz Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

2 READING QUIZ Answers: 1. C 2. A
1. Frames and machines are different as compared to trusses since they have ___________. A) Only two-force members B) Only multiforce members C) At least one multiforce member D) At least one two-force member 2. Forces common to any two contacting members act with _______ on the other member. A) Equal magnitudes but opposite sense B) Equal magnitudes and the same sense C) Different magnitudes but opposite sense D) Different magnitudes but the same sense Answers: 1. C 2. A Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

3 Frames are commonly used to support various external loads.
APPLICATIONS Frames are commonly used to support various external loads. How is a frame different than a truss? To be able to design a frame, you need to determine the forces at the joints and supports. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

4 APPLICATIONS (continued)
“Machines,” like those above, are used in a variety of applications. How are they different from trusses and frames? How can you determine the loads at the joints and supports? These forces and moments are required when designing the machine members. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

5 FRAMES AND MACHINES: DEFINITIONS
Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members). Frame Machine Frames are generally stationary and support external loads. Machines contain moving parts and are designed to alter the effect of forces. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

6 STEPS FOR ANALYZING A FRAME OR MACHINE
1. Draw a FBD of the frame or machine and its members, as necessary. Hints: a) Identify any two-force members, b) forces on contacting surfaces (usually between a pin and a member) are equal and opposite, and, c) for a joint with more than two members or an external force, it is advisable to draw a FBD of the pin. FAB 2. Develop a strategy to apply the equations of equilibrium to solve for the unknowns. Problems are going to be challenging since there are usually several unknowns. A lot of practice is needed to develop good strategies. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

7 Given: The frame supports an external load and moment as shown.
EXAMPLE Given: The frame supports an external load and moment as shown. Find: The horizontal and vertical components of the pin reactions at C and the magnitude of reaction at B. Plan: a) Draw FBDs of the frame member BC. Why pick this part of the frame? b) Apply the equations of equilibrium and solve for the unknowns at C and B. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

8 EXAMPLE FBD of member BC
CX CY B 45° FAB 400 N 1 m 2 m 800 N m FBD of member BC Please note that member AB is a two-force member. Equations of Equilibrium: +  MC = FAB sin45° (1) – FAB cos45° (3) N m (2) = 0 FAB = N Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

9 EXAMPLE FBD of member BC
CX CY B 45° FAB 400 N 1 m 2 m 800 N m FBD of member BC +  FY = – CY cos 45° – 400 = 0 CY = N +  FX = – CX sin 45° = 0 CX = N Now use the x and y direction Equations of Equilibrium: Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

10 CONCEPT QUIZ 1. The figures show a frame and its FBDs. If an additional couple moment is applied at C, then how will you change the FBD of member BC at B? A) No change, still just one force (FAB) at B. B) Will have two forces, BX and BY, at B. C) Will have two forces and a moment at B. D) Will add one moment at B. Answer: 1. A Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

11 CONCEPT QUIZ (continued)
2. The figures show a frame and its FBDs. If an additional force is applied at D, then how will you change the FBD of member BC at B? A) No change, still just one force (FAB) at B. B) Will have two forces, BX and BY, at B. C) Will have two forces and a moment at B. D) Will add one moment at B. Answer: 2. B Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

12 Given: A frame supports a 50-kg cylinder.
GROUP PROBLEM SOLVING Given: A frame supports a 50-kg cylinder. Find: The reactions that the pins exert on the frame at A and D. Plan: a) Draw a FBD of member ABC and another one for CD. b) Apply the equations of equilibrium to each FBD to solve for the six unknowns. Think about a strategy to easily solve for the unknowns. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

13 GROUP PROBLEM SOLVING (continued)
FBDs of members ABC and CD 50(9.81) N CY CX DX DY 0.7 m 1.6 m 1.2 m Applying E-of-E to member ABC: +  MA = CY (1.6) – 50 (9.81) (0.7) – 50 (9.81) (1.7) = 0 ; CY = N +  FY = AY – – 50 (9.81) – 50 (9.81) = 0 ; AY = 245 N +  FX = CX – AX = 0 ; CX = AX Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

14 GROUP PROBLEM SOLVING (continued)
FBDs of members ABC and CD 50(9.81) N CY CX DX DY 0.7 m 1.6 m 1.2 m Applying E-of-E to member CD: +  MD = CX (1.2) + 50 (9.81) (0.7) – 735.8(1.6) = 0 ; CX = 695 N +  FY = DY – (9.81) = 0 ; DY = 245 N +  FX = DX – = 0 ; DX = 695 N AX = CX = 695 N Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

15 ATTENTION QUIZ Answers: 1. A 2. C
1. When determining the reactions at joints A, B, and C, what is the minimum number of unknowns for solving this problem? A) 3 B) 4 C) 5 D) 6 2. For the above problem, imagine that you have drawn a FBD of member AB. What will be the easiest way to write an equation involving unknowns at B? A)  MC = 0 B)  MB = 0 C)  MA = 0 D)  FX = 0 Answers: 1. A 2. C Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6

16 End of the Lecture Let Learning Continue
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 6.6


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